Theorem fneq1i 5475

Description: Equality inference for function predicate with domain. (Contributed by Paul Chapman, 22-Jun-2011.)

Hypothesis

Ref	Expression
fneq1i.1	|- F = G

Assertion

Ref	Expression
fneq1i	|- ( F Fn A <-> G Fn A )

Proof of Theorem fneq1i

Step	Hyp	Ref	Expression
1		fneq1i.1	. 2 |- F = G
2		fneq1 5469	. 2 |- ( F = G -> ( F Fn A <-> G Fn A ) )
3	1, 2	ax-mp 5	1 |- ( F Fn A <-> G Fn A )

This proof depends on syntax axioms:   <-> wb 105   = wceq 1402   Fn wfn 5372

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-un 3224  df-in 3226  df-ss 3233  df-sn 3715  df-pr 3716  df-op 3718  df-br 4131  df-opab 4193  df-rel 4781  df-cnv 4782  df-co 4783  df-dm 4784  df-fun 5379  df-fn 5380

This theorem is used by:  fnunsn  5490  fnopabg  5507  f1oun  5659  f1oi  5679  f1osn  5681  ovid  6205  tfri1d  6606  frec2uzrand  10842  frec2uzf1od  10843  frecfzennn  10863  xnn0nnen  10874  prdsinvlem  14196  dfrelog  15961  edgstruct  16305  nninfsellemeqinf  17059